反三角函数证明

证明arcsin(2x²-1)-2arcsinx=-π/2
证明：两边取绝缓败正哪空弦：
左边＝sin[arcsin(2x²-1)-2arcsinx]
=sin[arcsin(2x²-1)]cos(2arcsinx)-cos[arcsin(2x²并颤-1)]sin(2arcsinx)
=(2x²-1)[1-2sin²(arcsinx)]-√[1-(2x²-1)²][2sin(arcsinx)cos(arcsinx)]
=(2x²-1)(1-2x²)-√[4x²(1-x²)](2x)√(1-x²)
=-(2x²-1)²-4x²(1-x²)=-(4x⁴-4x²+1)-(4x²-4x⁴)=-1
右边=sin(-π/2)=-1
故 arcsin(2x²-1)-2arcsinx=-π/2

设握磨y = arcsin(2x^2 - 1) - 2arcsinx
导数y' = {1/√[1-(2x^2-1)^2]} * 4x - 2/√(1-x^2)
= {1/√(1 - 4x^4 + 4x^2 - 1)} * 4x - 2/√(1-x^2)
= {1/√[4x^2(1-x^2)]} * 4x - 2/√(1-x^2)
= {1/[2x√(1-x^2)]} * 4x - 2/√(1-x^2)
= 2/√(1-x^2) - 2/√橡皮搭(1-x^2) = 0
由梁拿定理可知，y在其定义域[0,1]内为一常数C
将x=1带入y，由y = arcsin1 - 2arcsin1 = π/2 - π = -π/2
证明完毕

过程够详细了吧

先求定义域
-1≤2x^2-1≤1
-1≤x≤1
得-1≤x≤1
再看值域
由此可知，arc sinx∈[-π/2,π/2]
arc sin(2x^-1)∈[-π/2,π/2]
值域在区间[-3π/2,3π/2]上
对左边取正弦得
sin[arc sin(2x^-1)-2arc sinx]
=sin[arc sin(2x^2-1)]cos[2arcsinx]-cos[arc sin(2x^2-1)]sin[2arcsinx]
=(2x^2-1)*{[cos(arcsinx)]^2-[sin(arcsinx)]^2}-cos[arc sin(2x^2-1)]*2*sin[arcsinx]*cos[arcsinx]
=(2x^2-1)*(1-x^2-x^2)-√[1-(2x^2-1)^2]*2*x*√(1-x^2)
=-(2x^2-1)^2-2x√(1-x^2)*√4x^2(1-x^2)
1，如消当-1≤x<0时，该式得
=-(2x^2-1)^2+4x^2(1-x^2)
=-8x^4+8x^2-1
原式=arcsin(-8x^4+8x^2-1)
2，当0≤x≤1时，该式得
=-(2x^2-1)^2-4x^2(1-x^2)
=-1
且由于此渣宴知时值域无法取到3π/2，
所以原式=arcsin(-1)=-π/2
这祥模题很可能是少了个条件，x>0